A 8 Omega resistor is connected to a battery | vedclass

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(C) The terminal voltage $V$ of a battery is given by the relation $V = E - Ir$,where $E$ is the emf,$I$ is the current,and $r$ is the internal resist

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$10.25$

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(C) The terminal voltage $V$ of a battery is given by the relation $V = E - Ir$,where $E$ is the emf,$I$ is the current,and $r$ is the internal resistance.Given:External resistance $R = 8 \ \Omega$Internal resistance $r = 0.2 \ \Omega$Terminal voltage $V = 10 \ V$The current $I$ flowing through the circuit is given by Ohm's law applied to the external resistor:$I = \frac{V}{R} = \frac{10 \ V}{8 \ \Omega} = 1.25 \ A$Now,substituting the values into the terminal voltage equation:$V = E - Ir$$10 = E - (1.25 \ A 	imes 0.2 \ \Omega)$$10 = E - 0.25 \ V$$E = 10 + 0.25 = 10.25 \ V$Therefore,the emf of the battery is $10.25 \ V$.

$A$ $8 \Omega$ resistor is connected to a battery that has an internal resistance of $0.2 \Omega$. If the voltage across the battery (the terminal voltage) is $10 \ V$,then the emf of the battery is (in $V$)

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